I find printers from iPad setting print can find my Brother printer.
But when I try the code I get empty device list and I don't know why.
I am not familiar with Swift. I just try the sample code from Official documentation.
https://support.brother.com/g/s/es/htmldoc/mobilesdk/guide/discover-printer.html
Here is my code:
func getPrinter() {
let printerManager = BRPtouchNetworkManager()
printerManager.setPrinterName("Brother QL-720NW")
printerManager.startSearch(5)
printerManager.getPrinterNetInfo()
print("start")
let testFind = YourClass()
print("1")
testFind.startSearchWiFiPrinter()
testFind.didFinishSearch(printerManager)
print("2")
}
class YourClass: NSObject, BRPtouchNetworkDelegate {
private var networkManager: BRPtouchNetworkManager?
func startSearchWiFiPrinter() {
print("3")
let manager = BRPtouchNetworkManager()
manager.delegate = self
manager.startSearch(5)
self.networkManager = manager
}
// BRPtouchNetworkDelegate
func didFinishSearch(_ sender: Any!) {
print("4")
guard let manager = sender as? BRPtouchNetworkManager else {
print("5")
return
}
guard let devices = manager.getPrinterNetInfo() else {
print("6")
return
}
print(devices)
print("7")
for deviceInfo in devices {
print("8")
if let deviceInfo = deviceInfo as? BRPtouchDeviceInfo {
print("Model: \(deviceInfo.strModelName), IP Address: \(deviceInfo.strIPAddress)")
}
}
}
}
I call my function getPrinter() and here is my print log:
The SDK documentation gives you an example implementation of two methods:
func startSearchWiFiPrinter() {}
and
func didFinishSearch(_ sender: Any!) {}
In the class you want to execute the search you must implement these. You also need to declare the class to attend the protocol BRPtouchNetworkDelegate. The last thing is to have a property to be able to hold the Network manager (Which is done in the line: private var networkManager: BRPtouchNetworkManager?)
However, you are not supposed to call the "didFinishSearch" method by yourself. When you call startSearchWiFiPrinter, the search begins, and the BRPtouchNetworkManager instance itself calls the didFinishSearch method. It is capable of doing so because you set the delegate in the line: manager.delegate = self.
You should not need 2 classes for this. You should not use 2 instances of BRPtouchNetworkManager either.
Try this. Remember the number you put as an argument to startSearchWiFiPrinter means how long in seconds the search will be.
class EXAMPLEClass: NSObject, BRPtouchNetworkDelegate {
private var networkManager: BRPtouchNetworkManager?
func getPrinter() {
self.startSearchWiFiPrinter()
}
func startSearchWiFiPrinter() {
let manager = BRPtouchNetworkManager()
manager.delegate = self
manager.setPrinterName("Brother QL-720NW")
manager.startSearch(5)
self.networkManager = manager
}
// BRPtouchNetworkDelegate
func didFinishSearch(_ sender: Any!) {
print("4")
guard let manager = sender as? BRPtouchNetworkManager else {
print("5")
return
}
guard let devices = manager.getPrinterNetInfo() else {
print("6")
return
}
print(devices)
print("7")
for deviceInfo in devices {
print("8")
if let deviceInfo = deviceInfo as? BRPtouchDeviceInfo {
print("Model: \(deviceInfo.strModelName), IP Address: \(deviceInfo.strIPAddress)")
}
}
}
}
I want to check if the wifi is off then show alert to the user to check his/her connectivity.
I find code like this but it checks if there is an internet connection, not checking if the wifi is on or off:
func isConnectionAvailble()->Bool{
var rechability = SCNetworkReachabilityCreateWithName(kCFAllocatorDefault, "www.apple.com").takeRetainedValue()
var flags : SCNetworkReachabilityFlags = 0
if SCNetworkReachabilityGetFlags(rechability, &flags) == 0
{
return false
}
let isReachable = (flags & UInt32(kSCNetworkFlagsReachable)) != 0
let needsConnection = (flags & UInt32(kSCNetworkFlagsConnectionRequired)) != 0
return (isReachable && !needsConnection)
}
You can't.
With Apple's reachability class, you can distinguish three things according to the NetworkStatus struct:
typedef enum : NSInteger {
NotReachable = 0, // 1
ReachableViaWiFi, // 2
ReachableViaWWAN // 3
} NetworkStatus;
You have neither WiFi nor mobile data connection.
You have a WiFi connection, but you may or may not have a mobile data connection.
You have a mobile data connection, but no WiFi connection.
You can't check whether WiFi is turned off, or whether WiFi is turned on but there is no WiFi network nearby, or whether Airplane mode has been turned on.
For mobile data, you can use the telephony class to find whether your device is capable of mobile data connections (iPhone and not iPad, and SIM card plugged in), and you can detect whether mobile data is disabled in the preferences of your application.
Found the following, which was really helpful for me (found on the Apple Developer Forums). The below code works with Swift 4.
func fetchSSIDInfo() -> String {
var currentSSID = ""
if let interfaces:CFArray = CNCopySupportedInterfaces() {
for i in 0..<CFArrayGetCount(interfaces){
let interfaceName: UnsafeRawPointer = CFArrayGetValueAtIndex(interfaces, i)
let rec = unsafeBitCast(interfaceName, to: AnyObject.self)
let unsafeInterfaceData = CNCopyCurrentNetworkInfo("\(rec)" as CFString)
if unsafeInterfaceData != nil {
let interfaceData = unsafeInterfaceData! as Dictionary!
for dictData in interfaceData! {
if dictData.key as! String == "SSID" {
currentSSID = dictData.value as! String
}
}
}
}
}
return currentSSID
}
You can then check if a device is connected to Wi-Fi by the following:
if fetchSSIDInfo() != nil {
/* Wi-Fi is Connected */
}
Not perfect, but if the device is not connected to a Wi-Fi Network, you could then ask the user to connect to a Wi-Fi Network:
let wifiNotifcation = UIAlertController(title: "Please Connect to Wi-Fi", message: "Please connect to your standard Wi-Fi Network", preferredStyle: .alert)
wifiNotifcation.addAction(UIAlertAction(title: "Open Wi-Fi", style: .default, handler: { (nil) in
let url = URL(string: "App-Prefs:root=WIFI")
if UIApplication.shared.canOpenURL(url!){
UIApplication.shared.openURL(url!)
self.navigationController?.popViewController(animated: false)
}
}))
self.present(wifiNotifcation, animated: true, completion: nil)
Tested with swift 4 and swift 5
let nwPathMonitor = NWPathMonitor()
nwPathMonitor.pathUpdateHandler = { path in
if path.usesInterfaceType(.wifi) {
print("Path is Wi-Fi")
} else if path.usesInterfaceType(.cellular) {
print("Path is Cellular")
} else if path.usesInterfaceType(.wiredEthernet) {
print("Path is Wired Ethernet")
} else if path.usesInterfaceType(.loopback) {
print("Path is Loopback")
} else if path.usesInterfaceType(.other) {
print("Path is other")
}
}
nwPathMonitor.start(queue: .main)
As already #abba_de_bo mentioned: you could fetch the current SSID and check if it's set or nil.
This is the answer Apple's Eskimo gave to this question:
The trick with using CF-based APIs from Swift is to get the data into ‘Swift space’ as quickly as possible.
func currentSSIDs() -> [String] {
guard let interfaceNames = CNCopySupportedInterfaces() as? [String] else {
return []
}
return interfaceNames.flatMap { name in
guard let info = CNCopyCurrentNetworkInfo(name as CFString) as? [String:AnyObject] else {
return nil
}
guard let ssid = info[kCNNetworkInfoKeySSID as String] as? String else {
return nil
}
return ssid
}
}
Note that this returns an array of names; how you handle the non-standard cases (no elements, more than one element) is up to you.
Make sure you import SystemConfiguration.CaptiveNetwork. Otherwise the build will fail with on of those error messages:
Use of unresolved identifier 'CNCopySupportedInterfaces'
Use of unresolved identifier 'CNCopyCurrentNetworkInfo'
Use of unresolved identifier 'kCNNetworkInfoKeySSID'
You can take a look at the official Apple sample for Reachability:
https://developer.apple.com/library/content/samplecode/Reachability/Introduction/Intro.html
var netStatus = reachability.currentReachabilityStatus()
var connectionRequired = reachability.connectionRequired()
var statusString = ""
switch netStatus {
case NotReachable:
break
case ReachableViaWWAN:
//DATA
break
case ReachableViaWiFi:
//WIFI
break
}
You can use this method to check:
First you import this framework:
import SystemConfiguration.CaptiveNetwork
func isWifiEnabled() -> Bool {
var hasWiFiNetwork: Bool = false
let interfaces: NSArray = CFBridgingRetain(CNCopySupportedInterfaces()) as! NSArray
for interface in interfaces {
// let networkInfo = (CFBridgingRetain(CNCopyCurrentNetworkInfo(((interface) as! CFString))) as! NSDictionary)
let networkInfo: [AnyHashable: Any]? = CFBridgingRetain(CNCopyCurrentNetworkInfo(((interface) as! CFString))) as? [AnyHashable : Any]
if (networkInfo != nil) {
hasWiFiNetwork = true
break
}
}
return hasWiFiNetwork;
}
Trying to get the SSID of current device. I have found plenty of examples on how to do it however I am struggling with getting the CNCopySupportedInterfaces to autocomplete. I have 'import SystemConfiguration' at the top of my swift file but no success. Can't seem to figure out what I am doing wrong.
iOS 12
You must enable Access WiFi Information from capabilities.
Important
To use this function in iOS 12 and later, enable the Access WiFi Information capability for your app in Xcode. When you enable this capability, Xcode automatically adds the Access WiFi Information entitlement to your entitlements file and App ID. Documentation link
You need: import SystemConfiguration.CaptiveNetwork
Underneath the covers, CaptiveNetwork is a C header file (.h) that is within the SystemConfiguration framework:
/Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS.sdk/System/Library/Frameworks/SystemConfiguration.framework/Headers/CaptiveNetwork.h
If you know Objective-C, this goes into more depth:
iPhone get SSID without private library
You have to use the awkward syntax to bridge from any pure C API, so the following is required:
for interface in CNCopySupportedInterfaces().takeRetainedValue() as! [String] {
println("Looking up SSID info for \(interface)") // en0
let SSIDDict = CNCopyCurrentNetworkInfo(interface).takeRetainedValue() as! [String : AnyObject]
for d in SSIDDict.keys {
println("\(d): \(SSIDDict[d]!)")
}
}
ADDENDUM FOR SWIFT 2.2 and 3.0
The CFxxx datatypes are now bridged to native Objective-C runtime, eliminating the head-scratching retain calls. However, nullable pointers give rise to Optionals, so things don't get any shorter. At least, it's fairly clear what's going on, plus the nil helps us identify the simulator. The other answer uses an awful lot of bit-casting and unsafe operations which seems non-Swiftian, so I offer this.
func getInterfaces() -> Bool {
guard let unwrappedCFArrayInterfaces = CNCopySupportedInterfaces() else {
print("this must be a simulator, no interfaces found")
return false
}
guard let swiftInterfaces = (unwrappedCFArrayInterfaces as NSArray) as? [String] else {
print("System error: did not come back as array of Strings")
return false
}
for interface in swiftInterfaces {
print("Looking up SSID info for \(interface)") // en0
guard let unwrappedCFDictionaryForInterface = CNCopyCurrentNetworkInfo(interface) else {
print("System error: \(interface) has no information")
return false
}
guard let SSIDDict = (unwrappedCFDictionaryForInterface as NSDictionary) as? [String: AnyObject] else {
print("System error: interface information is not a string-keyed dictionary")
return false
}
for d in SSIDDict.keys {
print("\(d): \(SSIDDict[d]!)")
}
}
return true
}
Output on success:
SSIDDATA: <57696c6d 79>
BSSID: 12:34:56:78:9a:bc
SSID: YourSSIDHere
In Swift 2.0 / iOS 9 the API CaptiveNetwork is (nearly) gone or depreciated. I contacted Apple regarding this problem and I thought we could (or should) use the NEHotspotHelper instead. I got a respond from Apple today: One should continue to use CaptiveNetwork and the two relevant APIs (even tough there marked depreciated):
CNCopySupportedInterfaces
CNCopyCurrentNetworkInfo
The user braime posted an updated code-snippet for this problem on Ray Wenderlich forums:
let interfaces:CFArray! = CNCopySupportedInterfaces()
for i in 0..<CFArrayGetCount(interfaces){
let interfaceName: UnsafePointer<Void>
= CFArrayGetValueAtIndex(interfaces, i)
let rec = unsafeBitCast(interfaceName, AnyObject.self)
let unsafeInterfaceData = CNCopyCurrentNetworkInfo("\(rec)")
if unsafeInterfaceData != nil {
let interfaceData = unsafeInterfaceData! as Dictionary!
currentSSID = interfaceData["SSID"] as! String
} else {
currentSSID = ""
}
}
Works perfect for me.
Swift:
import SystemConfiguration.CaptiveNetwork
func currentSSIDs() -> [String] {
guard let interfaceNames = CNCopySupportedInterfaces() as? [String] else {
return []
}
return interfaceNames.flatMap { name in
guard let info = CNCopyCurrentNetworkInfo(name as CFString) as? [String:AnyObject] else {
return nil
}
guard let ssid = info[kCNNetworkInfoKeySSID as String] as? String else {
return nil
}
return ssid
}
}
Then print(currentSSIDs()), not working on simulator, only real devices.
Taken from https://forums.developer.apple.com/thread/50302
func getInterfaces() -> String? {
var ssid: String?
if let interfaces = CNCopySupportedInterfaces() as NSArray? {
for interface in interfaces {
if let interfaceInfo = CNCopyCurrentNetworkInfo(interface as! CFString) as NSDictionary? {
ssid = interfaceInfo[kCNNetworkInfoKeySSID as String] as? String
break
}
}
}
return ssid
}
In iOS 12 and up you will need to enable the Access WiFi Information capability for your app in order to get the ssid
I want to get all the WiFi networks available in a region and their SSID value. But the problem is how to get the SSID of all the WiFi network available even if I am not connected to one.
iOS 12
You must enable Access WiFi Information from capabilities.
Important
To use this function in iOS 12 and later, enable the Access WiFi Information capability for your app in Xcode. When you enable this capability, Xcode automatically adds the Access WiFi Information entitlement to your entitlements file and App ID. Documentation link
First;
import SystemConfiguration.CaptiveNetwork
Then;
func getInterfaces() -> Bool {
guard let unwrappedCFArrayInterfaces = CNCopySupportedInterfaces() else {
print("this must be a simulator, no interfaces found")
return false
}
guard let swiftInterfaces = (unwrappedCFArrayInterfaces as NSArray) as? [String] else {
print("System error: did not come back as array of Strings")
return false
}
for interface in swiftInterfaces {
print("Looking up SSID info for \(interface)") // en0
guard let unwrappedCFDictionaryForInterface = CNCopyCurrentNetworkInfo(interface) else {
print("System error: \(interface) has no information")
return false
}
guard let SSIDDict = (unwrappedCFDictionaryForInterface as NSDictionary) as? [String: AnyObject] else {
print("System error: interface information is not a string-keyed dictionary")
return false
}
for d in SSIDDict.keys {
print("\(d): \(SSIDDict[d]!)")
}
}
return true
}
Here my class that prints the WIFI network name
import UIKit
import Foundation
import SystemConfiguration.CaptiveNetwork
class FirstView: UIViewController
{
#IBOutlet weak var label: UILabel!
override func viewDidLoad()
{
super.viewDidLoad()
let ssid = self.getWiFiName()
print("SSID: \(ssid)")
}
func getWiFiName() -> String? {
var ssid: String?
if let interfaces = CNCopySupportedInterfaces() as NSArray? {
for interface in interfaces {
if let interfaceInfo = CNCopyCurrentNetworkInfo(interface as! CFString) as NSDictionary? {
ssid = interfaceInfo[kCNNetworkInfoKeySSID as String] as? String
break
}
}
}
return ssid
}
}
Yes it is possible to list all nearby WiFi networks.You need to complete a questionnaire at https://developer.apple.com/contact/network-extension, and then you can use NEHotspotHelper to return a list of hotspots. Technical Q&A https://developer.apple.com/library/archive/qa/qa1942/_index.html
How to achieve reflection in Swift Language?
How can I instantiate a class
[[NSClassFromString(#"Foo") alloc] init];
You must put #objc(SwiftClassName) above your swift class.
Like:
#objc(SubClass)
class SubClass: SuperClass {...}
This is the way I init derived UIViewController by class name
var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass()
More information is here
In iOS 9
var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass.init()
Less hacky solution here: https://stackoverflow.com/a/32265287/308315
Note that Swift classes are namespaced now so instead of "MyViewController" it'd be "AppName.MyViewController"
Deprecated since XCode6-beta 6/7
Solution developed using XCode6-beta 3
Thanks to the answer of Edwin Vermeer I was able to build something to instantiate Swift classes into an Obj-C class by doing this:
// swift file
// extend the NSObject class
extension NSObject {
// create a static method to get a swift class for a string name
class func swiftClassFromString(className: String) -> AnyClass! {
// get the project name
if var appName: String? = NSBundle.mainBundle().objectForInfoDictionaryKey("CFBundleName") as String? {
// generate the full name of your class (take a look into your "YourProject-swift.h" file)
let classStringName = "_TtC\(appName!.utf16count)\(appName)\(countElements(className))\(className)"
// return the class!
return NSClassFromString(classStringName)
}
return nil;
}
}
// obj-c file
#import "YourProject-Swift.h"
- (void)aMethod {
Class class = NSClassFromString(key);
if (!class)
class = [NSObject swiftClassFromString:(key)];
// do something with the class
}
EDIT
You can also do it in pure obj-c:
- (Class)swiftClassFromString:(NSString *)className {
NSString *appName = [[NSBundle mainBundle] objectForInfoDictionaryKey:#"CFBundleName"];
NSString *classStringName = [NSString stringWithFormat:#"_TtC%d%#%d%#", appName.length, appName, className.length, className];
return NSClassFromString(classStringName);
}
I hope this will help somebody !
UPDATE: Starting with beta 6 NSStringFromClass will return your bundle name plus class name separated by a dot. So it will be something like MyApp.MyClass
Swift classes will have a constructed internal name that is build up of the following parts:
It will start with _TtC,
followed by a number that is the length of your application name,
followed by your application name,
folowed by a number that is the length of your class name,
followed by your class name.
So your class name will be something like _TtC5MyApp7MyClass
You can get this name as a string by executing:
var classString = NSStringFromClass(self.dynamicType)
Update In Swift 3 this has changed to:
var classString = NSStringFromClass(type(of: self))
Using that string, you can create an instance of your Swift class by executing:
var anyobjectype : AnyObject.Type = NSClassFromString(classString)
var nsobjectype : NSObject.Type = anyobjectype as NSObject.Type
var rec: AnyObject = nsobjectype()
It's almost the same
func NSClassFromString(_ aClassName: String!) -> AnyClass!
Check this doc:
https://developer.apple.com/library/prerelease/ios/documentation/Cocoa/Reference/Foundation/Miscellaneous/Foundation_Functions/#//apple_ref/c/func/NSClassFromString
I was able to instantiate an object dynamically
var clazz: NSObject.Type = TestObject.self
var instance : NSObject = clazz()
if let testObject = instance as? TestObject {
println("yes!")
}
I haven't found a way to create AnyClass from a String (without using Obj-C). I think they don't want you to do that because it basically breaks the type system.
For swift2, I created a very simple extension to do this more quickly
https://github.com/damienromito/NSObject-FromClassName
extension NSObject {
class func fromClassName(className : String) -> NSObject {
let className = NSBundle.mainBundle().infoDictionary!["CFBundleName"] as! String + "." + className
let aClass = NSClassFromString(className) as! UIViewController.Type
return aClass.init()
}
}
In my case, i do this to load the ViewController I want:
override func viewDidLoad() {
super.viewDidLoad()
let controllers = ["SettingsViewController", "ProfileViewController", "PlayerViewController"]
self.presentController(controllers.firstObject as! String)
}
func presentController(controllerName : String){
let nav = UINavigationController(rootViewController: NSObject.fromClassName(controllerName) as! UIViewController )
nav.navigationBar.translucent = false
self.navigationController?.presentViewController(nav, animated: true, completion: nil)
}
This will get you the name of the class that you want to instantiate. Then you can use Edwins answer to instantiate a new object of your class.
As of beta 6 _stdlib_getTypeName gets the mangled type name of a variable. Paste this into an empty playground:
import Foundation
class PureSwiftClass {
}
var myvar0 = NSString() // Objective-C class
var myvar1 = PureSwiftClass()
var myvar2 = 42
var myvar3 = "Hans"
println( "TypeName0 = \(_stdlib_getTypeName(myvar0))")
println( "TypeName1 = \(_stdlib_getTypeName(myvar1))")
println( "TypeName2 = \(_stdlib_getTypeName(myvar2))")
println( "TypeName3 = \(_stdlib_getTypeName(myvar3))")
The output is:
TypeName0 = NSString
TypeName1 = _TtC13__lldb_expr_014PureSwiftClass
TypeName2 = _TtSi
TypeName3 = _TtSS
Ewan Swick's blog entry helps to decipher these strings: http://www.eswick.com/2014/06/inside-swift/
e.g. _TtSi stands for Swift's internal Int type.
In Swift 2.0 (tested in the Xcode 7.01) _20150930
let vcName = "HomeTableViewController"
let ns = NSBundle.mainBundle().infoDictionary!["CFBundleExecutable"] as! String
// Convert string to class
let anyobjecType: AnyObject.Type = NSClassFromString(ns + "." + vcName)!
if anyobjecType is UIViewController.Type {
// vc is instance
let vc = (anyobjecType as! UIViewController.Type).init()
print(vc)
}
xcode 7 beta 5:
class MyClass {
required init() { print("Hi!") }
}
if let classObject = NSClassFromString("YOURAPPNAME.MyClass") as? MyClass.Type {
let object = classObject.init()
}
string from class
let classString = NSStringFromClass(TestViewController.self)
or
let classString = NSStringFromClass(TestViewController.classForCoder())
init a UIViewController class from string:
let vcClass = NSClassFromString(classString) as! UIViewController.Type
let viewController = vcClass.init()
I am using this category for Swift 3:
//
// String+AnyClass.swift
// Adminer
//
// Created by Ondrej Rafaj on 14/07/2017.
// Copyright © 2017 manGoweb UK Ltd. All rights reserved.
//
import Foundation
extension String {
func convertToClass<T>() -> T.Type? {
return StringClassConverter<T>.convert(string: self)
}
}
class StringClassConverter<T> {
static func convert(string className: String) -> T.Type? {
guard let nameSpace = Bundle.main.infoDictionary?["CFBundleExecutable"] as? String else {
return nil
}
guard let aClass: T.Type = NSClassFromString("\(nameSpace).\(className)") as? T.Type else {
return nil
}
return aClass
}
}
The use would be:
func getViewController(fromString: String) -> UIViewController? {
guard let viewController: UIViewController.Type = "MyViewController".converToClass() else {
return nil
}
return viewController.init()
}
I think I'm right in saying that you can't, at least not with the current beta (2). Hopefully this is something that will change in future versions.
You can use NSClassFromString to get a variable of type AnyClass but there appears to be no way in Swift to instantiate it. You can use a bridge to Objective C and do it there or -- if it works in your case -- fall back to using a switch statement.
Apparently, it is not possible (anymore) to instantiate an object in Swift when the name of the class is only known at runtime. An Objective-C wrapper is possible for subclasses of NSObject.
At least you can instantiate an object of the same class as another object given at runtime without an Objective-C wrapper (using xCode Version 6.2 - 6C107a):
class Test : NSObject {}
var test1 = Test()
var test2 = test1.dynamicType.alloc()
In Swift 2.0 (tested in the beta2 of Xcode 7) it works like this:
protocol Init {
init()
}
var type = NSClassFromString(className) as? Init.Type
let obj = type!.init()
For sure the type coming from NSClassFromString have to implement this init protocol.
I expect it is clear, className is a String containing the Obj-C runtime name of the class which is by default NOT just "Foo", but this discussion is IMHO not the major topic of your question.
You need this protocol because be default all Swift classes don't implement an init method.
Looks like the correct incantation would be...
func newForName<T:NSObject>(p:String) -> T? {
var result:T? = nil
if let k:AnyClass = NSClassFromString(p) {
result = (k as! T).dynamicType.init()
}
return result
}
...where "p" stands for "packaged" – a distinct issue.
But the critical cast from AnyClass to T currently causes a compiler crash, so in the meantime one must bust initialization of k into a separate closure, which compiles fine.
I use different targets, and in this case the swift class is not found. You should replace CFBundleName with CFBundleExecutable. I also fixed the warnings:
- (Class)swiftClassFromString:(NSString *)className {
NSString *appName = [[NSBundle mainBundle] objectForInfoDictionaryKey:#"CFBundleExecutable"];
NSString *classStringName = [NSString stringWithFormat:#"_TtC%lu%#%lu%#", (unsigned long)appName.length, appName, (unsigned long)className.length, className];
return NSClassFromString(classStringName);
}
Isn't the solution as simple as this?
// Given the app/framework/module named 'MyApp'
let className = String(reflecting: MyClass.self)
// className = "MyApp.MyClass"
Also in Swift 2.0 (possibly before?) You can access the type directly with the dynamicType property
i.e.
class User {
required init() { // class must have an explicit required init()
}
var name: String = ""
}
let aUser = User()
aUser.name = "Tom"
print(aUser)
let bUser = aUser.dynamicType.init()
print(bUser)
Output
aUser: User = {
name = "Tom"
}
bUser: User = {
name = ""
}
Works for my use case
Try this.
let className: String = String(ControllerName.classForCoder())
print(className)
I have implemented like this,
if let ImplementationClass: NSObject.Type = NSClassFromString(className) as? NSObject.Type{
ImplementationClass.init()
}
Swift 5, easy to use, thanks to #Ondrej Rafaj's
Source code:
extension String {
fileprivate
func convertToClass<T>() -> T.Type? {
return StringClassConverter<T>.convert(string: self)
}
var controller: UIViewController?{
guard let viewController: UIViewController.Type = convertToClass() else {
return nil
}
return viewController.init()
}
}
class StringClassConverter<T> {
fileprivate
static func convert(string className: String) -> T.Type? {
guard let nameSpace = Bundle.main.infoDictionary?["CFBundleExecutable"] as? String, let aClass = NSClassFromString("\(nameSpace).\(className)") as? T.Type else {
return nil
}
return aClass
}
}
Call like this:
guard let ctrl = "ViewCtrl".controller else {
return
}
// ctrl do sth
A page jump example shown here, the hope can help you!
let vc:UIViewController = (NSClassFromString("SwiftAutoCellHeight."+type) as! UIViewController.Type).init()
self.navigationController?.pushViewController(vc, animated: true)
// Click the Table response
tableView.deselectRow(at: indexPath, animated: true)
let sectionModel = models[(indexPath as NSIndexPath).section]
var className = sectionModel.rowsTargetControlerNames[(indexPath as NSIndexPath).row]
className = "GTMRefreshDemo.\(className)"
if let cls = NSClassFromString(className) as? UIViewController.Type {
let dvc = cls.init()
self.navigationController?.pushViewController(dvc, animated: true)
}
Swift3+
extension String {
var `class`: AnyClass? {
guard
let dict = Bundle.main.infoDictionary,
var appName = dict["CFBundleName"] as? String
else { return nil }
appName.replacingOccurrences(of: " ", with: "_")
let className = appName + "." + self
return NSClassFromString(className)
}
}
Here is a good example:
class EPRocks {
#require init() { }
}
class EPAwesome : EPRocks {
func awesome() -> String { return "Yes"; }
}
var epawesome = EPAwesome.self();
print(epawesome.awesome);